Let r(cos O + i sin O) be a cube root of 125(cos 288 + i sin 288)
then
r^3(cos O + i sin O)^3 = 125(cos 288 + i sin 28)
so r^3 = 125 and cos 3O + i sin 3O = cos 288 + i sin 288
so r = 5 and 3O = 288 + 360p and O = 96 + 120p
so one cube root is 5 (cos 96 + i sin 96)
Im a little rusty at this stuff Its been a long time.
Im not sure of the other 2 roots
sorry cant help you any more
2*10^9 i'm pretty sure it would be d.
Given:
The expression is:
To find:
Part A: The expression using parentheses so that the expression equals 23.
Part B: The expression using parentheses so that the expression equals 3.
Solution:
Part A:
In option A,
[Using BODMAS]
In option B,
[Using BODMAS]
In option C,
In option D,
[Using BODMAS]
After the calculation, we have and .
Therefore, the correct options are B and D.
Part B: From part A, it is clear that
Therefore, the correct option is C.
96 divided by 6 is 16
96 divided by 8 is 12
96 divided by 12 is 8
96 divided by 16 is 6
96 divided by 32 is 3
Answer:
r=5.8
Step-by-step explanation:
Just do 17.4 divided by 3
And you get R
Hope this helps!
